Visual walkthrough — System with external forces — conditions for conservation

We build every symbol before we use it. If you have never seen an arrow used for a push, start at Step 1 and do not skip.

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Step 1 — What a single arrow means (momentum and force as arrows)

WHAT. Before any equation, we agree on the picture. A vector is just an arrow: its length tells you "how much," its direction tells you "which way." We draw two kinds of arrow for a single moving ball:

• $\vec{p}$ — the momentum arrow. It points the way the ball moves and is longer when the ball is heavier or faster. In symbols $\vec{p}=m\vec{v}$, where $m$ is the mass (a plain number, "how much stuff") and $\vec{v}$ is the velocity arrow (how fast, which way).
• $\vec{F}$ — the force arrow. It is the push or pull on the ball.

WHY these two. Newton discovered that force is not about where a thing is, but about how its momentum arrow changes. So we need momentum drawn as an arrow first, then force as the thing that stretches or turns it.

PICTURE. Look at the amber ball. Its cyan momentum arrow $\vec{p}$ points right. The white force arrow $\vec{F}$ nudges it. After a tiny time the momentum arrow has grown a little in the direction of $\vec{F}$ — that change is the small dashed arrow $d\vec{p}$.

Recall Why is

$\frac{d\vec{p}}{dt}$ the right tool and not $\Delta \vec{p}$? Because forces can vary from instant to instant. ::: A derivative gives the momentum change rate at each moment; a plain difference $\Delta\vec p$ would only give the total over a whole interval and hide how the force acted moment-to-moment.

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Step 2 — Drawing a system: many balls, one boundary

WHAT. A system is simply the group of objects we choose to circle. We draw a dashed loop — the boundary — around the balls we care about. Everything inside is "us"; everything outside is "the rest of the world."

We label the balls $1,2,3,\dots$ Ball number $i$ has mass $m_i$, velocity ${\vec{v}}_i$, and momentum ${\vec{p}}_i=m_i{\vec{v}}_i$.

WHY. The whole subject lives or dies on this loop. Whether a force is "internal" or "external" is not a property of the force — it depends entirely on where the dashed loop is drawn. So we draw it first, before judging any force.

PICTURE. Three amber balls sit inside a dashed cyan loop. The total momentum is the tip-to-tail sum of the three individual arrows — shown as the long white arrow $\vec{P}$.

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Step 3 — Sorting every force: internal twins vs outside pushes

WHAT. Now we look at every arrow pushing on ball $i$ and sort each one into a box:

• If the push comes from another ball $j$ inside the loop, call it an internal force ${\vec{f}}_{ij}$ ("force on $i$ from $j$").
• If the push comes from outside the loop (gravity, a hand, a wall), call it an external force ${\vec{F}}_i^{ext}$.

There is no third box. Every push has a source; that source is either inside the loop or outside it.

WHY. We are going to prove that the internal pushes secretly cancel and only the external ones survive. To do that we must first separate them cleanly on paper.

PICTURE. On ball $1$ we draw a cyan internal arrow (from ball $2$) and an amber external arrow (gravity, crossing the dashed boundary from outside). The forces are colour-coded by which box they belong to.

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Step 4 — Add up Newton's law over the whole system

WHAT. Write ${\vec{F}}_i=\frac{d{\vec{p}}_i}{dt}$ for each ball, then add all those equations together (sum over $i$).

WHY. We want a law about the whole group's momentum $\vec{P}$, not each ball separately. Adding the per-ball laws is exactly how the individual ${\vec{p}}_i$'s fuse into $\vec{P}$.

PICTURE. Three stacked rows, one per ball, each row an equation "force = rate of change of that ball's arrow." A big cyan brace on the right collects the three $\frac{d{\vec{p}}_i}{dt}$ into a single $\frac{d\vec{P}}{dt}$, because the derivative of a sum is the sum of derivatives.

${\sum }_i{\vec{F}}_i=\underset{\text{all internal}}{\underbrace{\sum _i\sum _{j\ne i}{\vec{f}}_{ij}}}+\underset{\text{all external}}{\underbrace{\sum _i{\vec{F}}_i^{ext}}}=\frac{d}{dt}{\sum }_i{\vec{p}}_i=\frac{d\vec{P}}{dt}$

• ${\sum }_i{\vec{F}}_i$ — total of all pushes on all balls.
• ${\sum }_i{\sum }_{j\ne i}{\vec{f}}_{ij}$ — the entire pile of internal pushes.
• ${\sum }_i{\vec{F}}_i^{ext}$ — the entire pile of external pushes.
• $\frac{d\vec{P}}{dt}$ — how the group's combined arrow changes. Why we can pull the sum inside the derivative: rate-of-change of (arrow 1 + arrow 2 + arrow 3) equals (rate of arrow 1) + (rate of arrow 2) + ..., so the sum and the $d/dt$ swap places freely.

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Step 5 — The magic cancellation (Newton's Third Law kills the internal pile)

WHAT. Take any pair of inside balls, say $1$ and $2$. Ball $2$ pushes ball $1$ with ${\vec{f}}_{12}$. Ball $1$ pushes back on ball $2$ with ${\vec{f}}_{21}$. Newton's Third Law says these are equal in length, opposite in direction: ${\vec{f}}_{12}=-{\vec{f}}_{21}$ So when both appear in the giant internal pile, they add to a zero arrow. Every internal push has a twin; every twin cancels it.

WHY. This is the reason internal forces — no matter how violent, like an explosion — cannot move the group's total arrow. They always come in equal-and-opposite pairs that erase each other in the sum.

PICTURE. Two amber balls; a cyan arrow from each pushing on the other, drawn perfectly equal and opposite. Below, the two arrows are laid tip-to-tail and collapse to a single point — the zero arrow $\vec{0}$.

${\sum }_i{\sum }_{j\ne i}{\vec{f}}_{ij}=\vec{0}$

• The double sum runs over every ordered pair, so both ${\vec{f}}_{12}$ and ${\vec{f}}_{21}$ are inside it.
• Grouped as twins $({\vec{f}}_{12}+{\vec{f}}_{21})$, each twin is a zero arrow, so the whole pile is $\vec{0}$.

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Step 6 — The master equation appears

WHAT. With the internal pile gone ($=\vec{0}$), only the external pile is left standing on the right: $\frac{d\vec{P}}{dt}={\sum }_i{\vec{F}}_i^{ext}={\vec{F}}_{ext}^{net}$

WHY. This is the payoff of all five previous steps. The group's total momentum arrow changes at exactly the rate of the net outside push — nothing internal ever enters.

PICTURE. The dashed loop again, but now only the amber external arrows survive, all summed into one bold amber arrow labelled ${\vec{F}}_{ext}^{net}$, sitting right beside the growing white arrow $\vec{P}$. Internal cyan arrows are faded to ghosts to show they dropped out.

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Step 7 — Edge case A: winning on one axis while losing on the other

WHAT. An arrow equation is secretly two equations: one for the horizontal ($x$) part, one for the vertical ($y$) part. They run independently: $\frac{d{P}_x}{dt}=F_{ext,x}^{net},\frac{d{P}_y}{dt}=F_{ext,y}^{net}$

So even if there is an outside push, if its horizontal part is zero, then $P_x$ is still conserved.

WHY. This is the cannon. Gravity points straight down — its horizontal shadow is zero. So sideways momentum of (cannon + ball) is safe even though up-down momentum is not.

PICTURE. A cannon fires a ball right; the recoil sends the cannon left. Gravity (amber, down) has no horizontal component (its $x$-shadow is a dot). The two horizontal momentum arrows are equal and opposite — their $x$-sum stays zero before and after.

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Step 8 — Edge case B: the impulsive approximation (short collisions)

WHAT. What actually changes momentum is impulse = force $\times$ time, written $\vec{J}=\int \vec{F}dt$ (the Impulse–Momentum Theorem). During a collision lasting a tiny $\Delta t$, a finite outside force like gravity delivers an almost-zero impulse: ${\vec{J}}_{ext}=\int {\vec{F}}_{ext}dt\approx {\vec{F}}_{ext}\Delta t\to \vec{0}(\Delta t\to 0)$

WHY. The contact force during a hit is enormous but brief; gravity is gentle. Their time is the same tiny sliver, so gravity's impulse is dwarfed. We treat $\vec{P}$ as conserved through the impact, then add gravity back for the slow flight.

PICTURE. Two force-vs-time graphs on the same axes: a tall thin cyan spike (contact force, huge but for $2$ ms) and a short flat amber line (gravity). The area under each is the impulse — the spike's amber-shaded area is tiny beside the... wait — the spike's area is large, gravity's area is a barely-visible sliver.

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Step 9 — Edge case C: the boundary decides everything (dropped ball)

WHAT. Same falling ball, two different loops:

• Loop around the ball only: gravity (from Earth) and the floor's push are outside the loop ⇒ external ⇒ $\vec{P}$ not conserved (it reverses on the bounce).
• Loop around ball + Earth: those same forces are now inside the loop ⇒ internal ⇒ $\vec{P}$ conserved (Earth recoils, imperceptibly).

WHY. "External" is never a property of a force alone — it is a property of the force relative to your loop. Redraw the loop, and the same force switches boxes.

PICTURE. Left panel: small dashed loop around the ball; gravity arrow pierces the boundary (external) — momentum arrow flips up after bounce. Right panel: big loop enclosing ball and Earth; the same force is now an internal twin pair — the combined arrow is unchanged. This is the same idea as Center of Mass Motion: with no external push, the centre of mass sails on unchanged.

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The one-picture summary

Everything above compressed into a single flow: force on each ball → split into internal + external → sum → internal twins cancel → only net external survives → if it's zero, $\vec{P}$ is frozen.

Recall Feynman retelling of the whole walkthrough

Picture a bag of marbles. Each marble has a little arrow showing how it's moving; add all the arrows tip-to-tail and you get one big arrow for the whole bag — that's the total momentum. Now the marbles bump each other inside the bag: every bump is a two-way shove, "I push you, you push me back exactly as hard the other way." Those cancel — so all the inside bumping in the world can't budge the bag's big arrow. The only thing that can budge it is a shove from outside the bag — a hand, gravity, a wall. So the big arrow stays frozen exactly when nothing outside is shoving. And three bonus tricks: (1) if the outside shove is only sideways, the forward part of the arrow is still frozen; (2) if a collision is over in a blink, gravity barely gets any time to shove, so the arrow is almost frozen right through the crash; (3) whether a force counts as "outside" depends entirely on where you drew the bag — draw the bag bigger and yesterday's outside shove becomes today's inside bump.

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Connections

• Parent topic — full note
• Newton's Third Law — the twin-cancellation in Step 5.
• Impulse–Momentum Theorem — the area-under-the-curve idea in Step 8.
• Center of Mass Motion — the same master equation restated in Step 9.
• Elastic vs Inelastic Collisions — the separate energy condition.
• Conservation of Linear Momentum — the special case ${\vec{F}}_{ext}^{net}=\vec{0}$.